The plane that contains the points (8,3,1), (2,6,3), and (4,6,2) has an equation of the form ax+by+cz=d. Find coefficients fr this equation, trying two different approaches to the problem. One method uses vectors, another does not.
The plane that contains the points (8,3,1), (2,6,3), and (4,6,2) has an equation of the form ax+by+cz=d. Find coefficients fr this equation, trying two different approaches to the problem. One method uses vectors, another does not.
Since the points $\displaystyle P_{1}, \;\ P_{2}, \;\ P_{3}$ lie in the plane, the vectors $\displaystyle \overline{P_{1}P_{2}}=(-6,3,2)$ and
$\displaystyle \overline{P_{1}P_{3}}=(-4,3,1)$ are parallel to the plane.
The cross product:
$\displaystyle P_{1}P_{2}\times{P_{1}P_{3}}=\begin{vmatrix}i&j&k\ \-6&3&-2\\-4&3&1\end{vmatrix}=-3i-2j-6k$
is normal to the plane.
Use this normal and the point (8,3,1):
$\displaystyle -3(x-8)-2(y-3)-6(z-1)=0$
And we get:
$\displaystyle -3x-2y-6z+36=0$
$\displaystyle \boxed{3x+2y+6z=36}$
Now, try the other method.